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A mixture of pulverized fuel ash and Portland cement to be used for grouting should have...

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with σ = 59.Let μ denote the true average compressive strength.

(a) What are the appropriate null and alternative hypotheses?


(b) Let X denote the sample average compressive strength for n = 11 randomly selected specimens. Consider the test procedure with test statistic X itself (not standardized). What is the probability distribution of the test statistic when H0 is true?


(c)If X = 1340, find the P-value. (Round your answer to four decimal places.)
P-value =  

(d)Should H0 be rejected using a significance level of 0.01?


(e) What is the probability distribution of the test statistic when μ = 1350?


(f)State the mean and standard deviation of the test statistic. (Round your standard deviation to three decimal places.)

mean        KN/m2
standard deviation        KN/m2


(g)For a test with α = 0.01, what is the probability that the mixture will be judged unsatisfactory when in fact μ = 1350 (a type II error)? (Round your answer to four decimal places.)


Homework Answers

Answer #1

b) Compressive strength is normally distributed with  standard deviation and mean

we take sample of size n= 11, Let X be the sample average of the compressive strength is normally distributed with   standard deviation and mean when H0 is true

c) X=1340

p-value = P(X>1340) = 0.0123

d) . there for we don't reject the null hypothesis

e)Let X be the sample average of the compressive strength is normally distributed with   standard deviation and mean

f) mean =1350

standard deviation =17.790

g)Type 2 error = P( accept H0 | H1 is true)

when

then the critical region is X> c, where is is determine such that P(X>c) =0.01

therefore c = 1341

then P(type 2 error ) =P(X<1341) =0.3065

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